# To fill a swimming pool two pipes are used. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find how long it would take for each pipe to fill the pool separately if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?

Answer

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Hint: First of all assume the time taken by a larger pipe to fill the pool as ‘T’ hours. From this, we will get the time for a smaller pipe as well. Then find the capacity of the pool filled by each of the pipes in one hour. Then use,

(Pool filled by the pipe of larger diameter in 4 hours) + (Pool filled by the pipe of smaller diameter in 9 hours) = \[\dfrac{1}{2}\] where ‘L’ is the total capacity of the pool.

Complete step-by-step answer:

Here we are given two pipes, one of larger diameter and other of smaller diameter which takes 4 hours and 9 hours respectively together to fill half of the pool. Also, we are given that pipe of the smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool. We have to find the time taken by each pipe to separately fill the pool.

Let us consider the total capacity of the pool to be ‘L’ liters.

Also, let us assume that a pipe of larger diameter takes ‘T’ hours to fill this pool.

Since this pipe takes ‘T’ hours to fill ‘L’ liters. Therefore, in 1 hour, it would fill \[=\dfrac{L}{T}\text{liters}\].

So, in 4 hours this pipe of larger diameter would fill \[=4\times \dfrac{L}{T}\text{liters}.....\left( i \right)\]

Now, we are also given that the pipe of the smaller diameter takes 10 more hours to fill the pool as compared to the pipe of the larger diameter. So, we get,

The time taken by a smaller pipe to fill the pool = (T + 10) hours.

Since this pipe takes ‘(T + 10)’ hours to fill ‘L’ liters, therefore in 1 hour, it would fill \[=\dfrac{L}{\left( T+10 \right)}\text{liters}\].

So, in 9 hours this pipe of smaller diameter would fill \[=9\times \dfrac{L}{\left( T+10 \right)}\text{liters}....\left( ii \right)\]

Now, we are given that pipe of larger diameter in 4 hours and pipe of smaller diameter in 9 hours together fills half of the pool, that is, \[\dfrac{L}{2}\text{liters}\]. So, we get,

(Pool filled by the pipe of larger diameter in 4 hours) + (Pool filled by the pipe of smaller diameter in 9 hours) = \[\dfrac{L}{2}\]

By substituting the values from equation (i) and (ii), we get,

\[\dfrac{4L}{T}+\dfrac{9L}{T+10}=\dfrac{L}{2}\]

By taking out ‘L’ common from both the sides and canceling it, we get,

\[\dfrac{4}{T}+\dfrac{9}{T+10}=\dfrac{1}{2}\]

By simplifying the above equation, we get,

\[\dfrac{4\left( T+10 \right)+9T}{T\left( T+10 \right)}=\dfrac{1}{2}\]

By cross multiplying the above equation, we get,

\[2\left( 4T+40+9T \right)={{T}^{2}}+10T\]

\[\Rightarrow 2\left( 13T+40 \right)={{T}^{2}}+10T\]

\[\Rightarrow 26T+80={{T}^{2}}+10T\]

By taking all the terms to one side, we get,

\[\Rightarrow {{T}^{2}}+10T-26T-80=0\]

Or, \[{{T}^{2}}-16T-80=0\]

Here, we can write \[16T=20T-4T\], so we get,

\[\Rightarrow {{T}^{2}}-20T+4T-80=0\]

We can also write the above equation as,

\[T\left( T-20 \right)+4\left( T-20 \right)=0\]

By taking out (T – 20) common, we get,

\[\left( T-20 \right)\left( T+4 \right)=0\]

We get, T = 20 and T = – 4

Since we know that the value of T can’t be negative, hence we get T = 20 hours.

Therefore, we get the total time taken by the pipe of larger diameter to fill the pool is equal to 20 hours.

Also, we get the total time taken by the pipe of smaller diameter to fill the pool is equal to (T + 10) = 30 hours.

Note: In these types of questions, students should always use the approach of the unitary method to solve the questions, that is, always calculate for 1 unit and they multiply it with the number of units you are asked for. Also, students can cross-check their answer by substituting the value of T = 20 hours back in the equation and checking if LHS = RHS or not.

(Pool filled by the pipe of larger diameter in 4 hours) + (Pool filled by the pipe of smaller diameter in 9 hours) = \[\dfrac{1}{2}\] where ‘L’ is the total capacity of the pool.

Complete step-by-step answer:

Here we are given two pipes, one of larger diameter and other of smaller diameter which takes 4 hours and 9 hours respectively together to fill half of the pool. Also, we are given that pipe of the smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool. We have to find the time taken by each pipe to separately fill the pool.

Let us consider the total capacity of the pool to be ‘L’ liters.

Also, let us assume that a pipe of larger diameter takes ‘T’ hours to fill this pool.

Since this pipe takes ‘T’ hours to fill ‘L’ liters. Therefore, in 1 hour, it would fill \[=\dfrac{L}{T}\text{liters}\].

So, in 4 hours this pipe of larger diameter would fill \[=4\times \dfrac{L}{T}\text{liters}.....\left( i \right)\]

Now, we are also given that the pipe of the smaller diameter takes 10 more hours to fill the pool as compared to the pipe of the larger diameter. So, we get,

The time taken by a smaller pipe to fill the pool = (T + 10) hours.

Since this pipe takes ‘(T + 10)’ hours to fill ‘L’ liters, therefore in 1 hour, it would fill \[=\dfrac{L}{\left( T+10 \right)}\text{liters}\].

So, in 9 hours this pipe of smaller diameter would fill \[=9\times \dfrac{L}{\left( T+10 \right)}\text{liters}....\left( ii \right)\]

Now, we are given that pipe of larger diameter in 4 hours and pipe of smaller diameter in 9 hours together fills half of the pool, that is, \[\dfrac{L}{2}\text{liters}\]. So, we get,

(Pool filled by the pipe of larger diameter in 4 hours) + (Pool filled by the pipe of smaller diameter in 9 hours) = \[\dfrac{L}{2}\]

By substituting the values from equation (i) and (ii), we get,

\[\dfrac{4L}{T}+\dfrac{9L}{T+10}=\dfrac{L}{2}\]

By taking out ‘L’ common from both the sides and canceling it, we get,

\[\dfrac{4}{T}+\dfrac{9}{T+10}=\dfrac{1}{2}\]

By simplifying the above equation, we get,

\[\dfrac{4\left( T+10 \right)+9T}{T\left( T+10 \right)}=\dfrac{1}{2}\]

By cross multiplying the above equation, we get,

\[2\left( 4T+40+9T \right)={{T}^{2}}+10T\]

\[\Rightarrow 2\left( 13T+40 \right)={{T}^{2}}+10T\]

\[\Rightarrow 26T+80={{T}^{2}}+10T\]

By taking all the terms to one side, we get,

\[\Rightarrow {{T}^{2}}+10T-26T-80=0\]

Or, \[{{T}^{2}}-16T-80=0\]

Here, we can write \[16T=20T-4T\], so we get,

\[\Rightarrow {{T}^{2}}-20T+4T-80=0\]

We can also write the above equation as,

\[T\left( T-20 \right)+4\left( T-20 \right)=0\]

By taking out (T – 20) common, we get,

\[\left( T-20 \right)\left( T+4 \right)=0\]

We get, T = 20 and T = – 4

Since we know that the value of T can’t be negative, hence we get T = 20 hours.

Therefore, we get the total time taken by the pipe of larger diameter to fill the pool is equal to 20 hours.

Also, we get the total time taken by the pipe of smaller diameter to fill the pool is equal to (T + 10) = 30 hours.

Note: In these types of questions, students should always use the approach of the unitary method to solve the questions, that is, always calculate for 1 unit and they multiply it with the number of units you are asked for. Also, students can cross-check their answer by substituting the value of T = 20 hours back in the equation and checking if LHS = RHS or not.

Last updated date: 21st Sep 2023

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